Given that the lines : (x-1)/2=(y+2)/3=z/5 and (x+2)/1=(y-1)/2=(z-α )…

The correct answer is Option (4) → -15

$\vec{v_1}=2\hat i+3\hat j+5\hat k$ || line 1

$\vec{v_2}=\hat i+2\hat j+2\hat k$ || line 2

points on line 1: $\vec{α_1}=\hat i-2\hat j$

line 2: $\vec{α_2}=-2\hat i+\hat j+α\hat k$

so for coplanarity

 $[\vec{α_2}-\vec{α_1}\,\,\,\vec{v_1}\vec{v_2}]=0$

$⇒\begin{vmatrix}-3&3&α\\2&3&5\\1&2&2\end{vmatrix}=0$

$α+15=0$

$⇒α=-15$